Answer:
a) 0.0498
b) 0.1489
c) 0.1818
Step-by-step explanation:
Given:
Number of telephones = 6+6+6= 18
6 cellular, 6 cordless, and 6 corded.
a) Probability that all the cordless phones are among the first twelve to be serviced:
12 are selected from 18 telephones, possible number of ways of selection = ¹⁸C₁₂
Then 6 cordless telephones are serviced, the remaining telephones are: 12 - 6 = 6.
The possible ways of selecting thr remaining 6 telephones = ¹²C₆
Probability of servicing all cordless phones among the first twelve:
= (⁶C₆) (⁶C₁₂) / (¹⁸C₁₂)
[tex] = \frac{1 * 924}{18564} [/tex]
[tex] = 0.0498 [/tex]
b) Probability that after servicing twelve of these phones, phones of only two of the three types remain to be serviced:
Here,
One type must be serviced first
The 6 remaining to be serviced can be a combination of the remaining two types.
Since there a 3 ways to select one type to be serviced, the probability will be:
= 3 [(⁶C₁)(⁶C₅) + (⁶C₂)(⁶C₄) + (⁶C₃)(⁶C₃) + (⁶C₄)(⁶C₂) + (⁶C₅)(⁶C₁)] / ¹⁸C₁₂
[tex] = \frac{3 * [(6)(6) + (15)(15) + (20)(20) + (15)(15) + (6)(6)]}{18564}[/tex]
[tex] = \frac{2766}{18564} [/tex]
[tex] = 0.1489 [/tex]
c) probability that two phones of each type are among the first six:
(⁶C₂)³/¹⁸C₆
[tex] \frac{3375}{18564}[/tex]
[tex] =0.1818 [/tex]
